Problem: Suppose we have a plane $P$ defined by the transformation $T$ for $-3 < u < 1$ and $0 < v < \sqrt{2}$. $T(u, v) = (2u + 3v, u - 5v, u + v)$ What is the surface area of $P$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4\sqrt{103}$ (Choice B) B $4\sqrt{206}$ (Choice C) C $8\sqrt{103}$ (Choice D) D $8\sqrt{206}$
Solution: Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_0^{\sqrt{2}} \int_{-3}^1 |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = \sqrt{206}$ [Calculation] The final step is to evaluate the double integral. $\begin{aligned} A &= \int_0^{\sqrt{2}} \int_{-3}^1 |T_u \times T_v| \, du \, dv \\ \\ &= \int_0^{\sqrt{2}} \int_{-3}^1 \sqrt{206} \, du \, dv \\ \\ &= \int_0^{\sqrt{2}} 4\sqrt{206} \, dv \\ \\ &= 4 \sqrt{2} \sqrt{206} \\ \\ &= 8 \sqrt{103} \end{aligned}$ In conclusion, the surface area of $P$ is $8 \sqrt{103}$.